Young’s modulus of a uniform cross section rod, linearly varies from $Y_0\left(endA\right)to2Y_0\left(endB\right)$. Two equal forces are acting at ends of rod (area of cross-section =A, length = l) 

Stress is $\frac{F}{A}$,  $\therefore$ it will be same at all points.

Young’s modulus on element dx will be  $Y_0+\frac{Y_0}{L}x$
Let extension in dx be dy
then $\frac{F}{A}=(Y_0+\frac{Y_{0}x}{L})\frac{dy}{dx}$
$\underset{0}{\overset{L}{\int }}\frac{FL}{A}\frac{dx}{(Y_{0}L+Y_{0}x)}=\underset{0}{\overset{\Delta L}{\int }}dy$

$\Delta L:$ total alongation in rod.